人工智能—Python实现线性回归

1、概述

（1）人工智能学习

（2）机器学习

（3）有监督学习

（4）线性回归

2、线性回归

（1）实现步骤

• 根据随机初始化的 w x b 和 y 来计算 loss
• 根据当前的 w x b 和 y 的值来计算梯度
• 更新梯度，循环将新的 w′ 和 b′ 复赋给 w 和 b ，最终得到一个最优的 w′ 和 b′ 作为方程最终的

（2）数学表达式

3、代码实现（Python）

（1）机器学习库（sklearn.linear_model）

代码：

from sklearn import linear_model
from sklearn.linear_model import LinearRegression
import matplotlib.pyplot as plt#用于作图
from pylab import *
mpl.rcParams['font.sans-serif'] = ['SimHei']
mpl.rcParams['axes.unicode_minus'] = False
import numpy as np#用于创建向量
 
 
reg=linear_model.LinearRegression(fit_intercept=True,normalize=False)
x=[[32.50235],[53.4268],[61.53036],[47.47564],[59.81321],[55.14219],[52.14219],[39.29957],
 [48.10504],[52.55001],[45.41873],[54.35163],[44.16405],[58.16847],[56.72721]]
y=[31.70701,68.7776,62.56238,71.54663,87.23093,78.21152,79.64197,59.17149,75.33124,71.30088,55.16568,82.47885,62.00892
,75.39287,81.43619]
reg.fit(x,y)
k=reg.coef_#获取斜率w1,w2,w3,...,wn
b=reg.intercept_#获取截距w0
x0=np.arange(30,60,0.2)
y0=k*x0+b
print("k={0},b={1}".format(k,b))
plt.scatter(x,y)
plt.plot(x0,y0,label='LinearRegression')
plt.xlabel('X')
plt.ylabel('Y')
plt.legend()
plt.show()

结果：

k=[1.36695374],b=0.13079331831460195

（2）Python详细实现（方法1）

代码：

#方法1
import numpy as np
import matplotlib.pyplot as plt
from pylab import *
mpl.rcParams['font.sans-serif'] = ['SimHei']
mpl.rcParams['axes.unicode_minus'] = False
#数据生成
data = []
for i in range(100):
    x = np.random.uniform(3., 12.)
    # mean=0, std=1
    eps = np.random.normal(0., 1)
    y = 1.677 * x + 0.039 + eps
    data.append([x, y])
 
data = np.array(data)
 
#统计误差
# y = wx + b
def compute_error_for_line_given_points(b, w, points):
    totalError = 0
    for i in range(0, len(points)):
        x = points[i, 0]
        y = points[i, 1]
        # computer mean-squared-error
        totalError += (y - (w * x + b)) ** 2
    # average loss for each point
    return totalError / float(len(points))
 
 
#计算梯度
def step_gradient(b_current, w_current, points, learningRate):
    b_gradient = 0
    w_gradient = 0
    N = float(len(points))
    for i in range(0, len(points)):
        x = points[i, 0]
        y = points[i, 1]
        # grad_b = 2(wx+b-y)
        b_gradient += (2/N) * ((w_current * x + b_current) - y)
        # grad_w = 2(wx+b-y)*x
        w_gradient += (2/N) * x * ((w_current * x + b_current) - y)
    # update w'
    new_b = b_current - (learningRate * b_gradient)
    new_w = w_current - (learningRate * w_gradient)
    return [new_b, new_w]
 
#迭代更新
def gradient_descent_runner(points, starting_b, starting_w, learning_rate, num_iterations):
    b = starting_b
    w = starting_w
    # update for several times
    for i in range(num_iterations):
        b, w = step_gradient(b, w, np.array(points), learning_rate)
    return [b, w]
 
 
def main():
 
    learning_rate = 0.0001
    initial_b = 0  # initial y-intercept guess
    initial_w = 0  # initial slope guess
    num_iterations = 1000
    print("迭代前 b = {0}, w = {1}, error = {2}"
          .format(initial_b, initial_w,
                  compute_error_for_line_given_points(initial_b, initial_w, data))
          )
    print("Running...")
    [b, w] = gradient_descent_runner(data, initial_b, initial_w, learning_rate, num_iterations)
    print("第 {0} 次迭代结果 b = {1}, w = {2}, error = {3}".
          format(num_iterations, b, w,
                 compute_error_for_line_given_points(b, w, data))
          )
    plt.plot(data[:,0],data[:,1], color='b', marker='+', linestyle='--',label='true')
    plt.plot(data[:,0],w*data[:,0]+b,color='r',label='predict')
    plt.xlabel('X')
    plt.ylabel('Y')
    plt.legend()
    plt.show()
 
 
if __name__ == '__main__':
    main()
 
 

 结果：

迭代前 ：b = 0, w = 0, error = 186.61000821356697
Running...
第 1000 次迭代结果：b = 0.20558501549252192, w = 1.6589067569038516, error = 0.9963685680112963

（3）Python详细实现（方法2）

代码：

#方法2
 
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib as mpl
mpl.rcParams["font.sans-serif"]=["SimHei"]
mpl.rcParams["axes.unicode_minus"]=False
 
 
# y = wx + b
#Import data
file=pd.read_csv("data.csv")
 
def compute_error_for_line_given(b, w):
    totalError = np.sum((file['y']-(w*file['x']+b))**2)
    return np.mean(totalError)
 
def step_gradient(b_current, w_current,  learningRate):
    b_gradient = 0
    w_gradient = 0
    N = float(len(file['x']))
    for i in range (0,len(file['x'])):
        # grad_b = 2(wx+b-y)
        b_gradient += (2 / N) * ((w_current * file['x'] + b_current) - file['y'])
        # grad_w = 2(wx+b-y)*x
        w_gradient += (2 / N) * file['x'] * ((w_current * file['x'] + b_current) - file['x'])
    # update w'
    new_b = b_current - (learningRate * b_gradient)
    new_w = w_current - (learningRate * w_gradient)
    return [new_b, new_w]
 
 
def gradient_descent_runner( starting_b, starting_w, learning_rate, num_iterations):
    b = starting_b
    w = starting_w
    # update for several times
    for i in range(num_iterations):
        b, w = step_gradient(b, w,  learning_rate)
    return [b, w]
 
 
def main():
    learning_rate = 0.0001
    initial_b = 0  # initial y-intercept guess
    initial_w = 0  # initial slope guess
    num_iterations = 100
    print("Starting gradient descent at b = {0}, w = {1}, error = {2}"
          .format(initial_b, initial_w,
                  compute_error_for_line_given(initial_b, initial_w))
          )
    print("Running...")
    [b, w] = gradient_descent_runner(initial_b, initial_w, learning_rate, num_iterations)
    print("After {0} iterations b = {1}, w = {2}, error = {3}".
          format(num_iterations, b, w,
                 compute_error_for_line_given(b, w))
          )
    plt.plot(file['x'],file['y'],'ro',label='线性回归')
    plt.xlabel('X')
    plt.ylabel('Y')
    plt.legend()
    plt.show()
 
 
 
 
if __name__ == '__main__':
    main()

结果：

Starting gradient descent at b = 0, w = 0, error = 75104.71822821398
Running...
After 100 iterations b = 0     0.014845
1     0.325621
2     0.036883
3     0.502265
4     0.564917
5     0.479366
6     0.568968
7     0.422619
8     0.565073
9     0.393907
10    0.216854
11    0.580750
12    0.379350
13    0.361574
14    0.511651
dtype: float64, w = 0     0.999520
1     0.994006
2     0.999405
3     0.989645
4     0.990683
5     0.991444
6     0.989282
7     0.989573
8     0.988498
9     0.992633
10    0.995329
11    0.989490
12    0.991617
13    0.993872
14    0.991116
dtype: float64, error = 6451.5510231710905

数据： 

（4）Python详细实现（方法3）

#方法3
 
import numpy as np
 
points = np.genfromtxt("data.csv", delimiter=",")
#从数据读入到返回需要两个迭代循环，第一个迭代将文件中每一行转化为一个字符串序列，
#第二个循环迭代对每个字符串序列指定合适的数据类型:
# y = wx + b
def compute_error_for_line_given_points(b, w, points):
    totalError = 0
    for i in range(0, len(points)):
        x = points[i, 0]
        y = points[i, 1]
        # computer mean-squared-error
        totalError += (y - (w * x + b)) ** 2
    # average loss for each point
    return totalError / float(len(points))
 
 
def step_gradient(b_current, w_current, points, learningRate):
    b_gradient = 0
    w_gradient = 0
    N = float(len(points))
    for i in range(0, len(points)):
        x = points[i, 0]
        y = points[i, 1]
        # grad_b = 2(wx+b-y)
        b_gradient += (2 / N) * ((w_current * x + b_current) - y)
        # grad_w = 2(wx+b-y)*x
        w_gradient += (2 / N) * x * ((w_current * x + b_current) - y)
    # update w'
    new_b = b_current - (learningRate * b_gradient)
    new_w = w_current - (learningRate * w_gradient)
    return [new_b, new_w]
 
 
def gradient_descent_runner(points, starting_b, starting_w, learning_rate, num_iterations):
    b = starting_b
    w = starting_w
    # update for several times
    for i in range(num_iterations):
        b, w = step_gradient(b, w, np.array(points), learning_rate)
    return [b, w]
 
 
def main():
    learning_rate = 0.0001
    initial_b = 0  # initial y-intercept guess
    initial_w = 0  # initial slope guess
    num_iterations = 1000
    print("Starting gradient descent at b = {0}, w = {1}, error = {2}"
          .format(initial_b, initial_w,
                  compute_error_for_line_given_points(initial_b, initial_w, points))
          )
    print("Running...")
    [b, w] = gradient_descent_runner(points, initial_b, initial_w, learning_rate, num_iterations)
    print("After {0} iterations b = {1}, w = {2}, error = {3}".
          format(num_iterations, b, w,
                 compute_error_for_line_given_points(b, w, points))
          )
 
 
if __name__ == '__main__':
    main()

4、案例——房屋与价格、尺寸

（1）代码

#1.导入包
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
from sklearn import linear_model
 
#2.加载训练数据，建立回归方程
# 取数据集(1)
datasets_X = []     #存放房屋面积
datasets_Y = []     #存放交易价格
fr = open('房价与房屋尺寸.csv','r')    #读取文件，r: 以只读方式打开文件，w: 打开一个文件只用于写入。
lines = fr.readlines()              #一次读取整个文件。
for line in lines:                  #逐行进行操作，循环遍历所有数据
    items = line.strip().split(',')    #去除数据文件中的逗号,strip()用于移除字符串头尾指定的字符（默认为空格或换行符）或字符序列。
                                       #split（‘ '）: 通过指定分隔符对字符串进行切片，如果参数 num 有指定值，则分隔 num+1 个子字符串。
    datasets_X.append(int(items[0]))   #将读取的数据转换为int型，并分别写入
    datasets_Y.append(int(items[1]))
 
length = len(datasets_X)              #求得datasets_X的长度，即为数据的总数
datasets_X = np.array(datasets_X).reshape([length,1])   #将datasets_X转化为数组，并变为1维，以符合线性回归拟合函数输入参数要求
datasets_Y = np.array(datasets_Y)                    #将datasets_Y转化为数组
 
#取数据集(2)
'''fr = pd.read_csv('房价与房屋尺寸.csv',encoding='utf-8')
datasets_X=fr['房屋面积']
datasets_Y=fr['交易价格']'''
 
minX = min(datasets_X)
maxX = max(datasets_X)
X = np.arange(minX,maxX).reshape([-1,1])        #以数据datasets_X的最大值和最小值为范围，建立等差数列，方便后续画图。
                                                #reshape([-1,1]），转换成１列，reshape([２,－1]）：转换成两行
linear = linear_model.LinearRegression()      #调用线性回归模块，建立回归方程，拟合数据
linear.fit(datasets_X, datasets_Y)
 
#3.斜率及截距
print('Coefficients:', linear.coef_)      #查看回归方程系数(k)
print('intercept:', linear.intercept_)    ##查看回归方程截距(b)
print('y={0}x+{1}'.format(linear.coef_,linear.intercept_)) #拟合线
 
# 4.图像中显示
plt.scatter(datasets_X, datasets_Y, color = 'red')
plt.plot(X, linear.predict(X), color = 'blue')
plt.xlabel('Area')
plt.ylabel('Price')
plt.show()

（2）结果

Coefficients: [0.14198749]
intercept: 53.43633899175563
y=[0.14198749]x+53.43633899175563

（3）数据

第一列是房屋面积，第二列是交易价格：

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